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Projectile Motion

Physics Unit 5 Lesson 1. Projectile Motion. Falling things, and rockets ‘n’ that…. Projectile Motion. If an object is launched at an initial angle of θ 0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component .

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Projectile Motion

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  1. Physics Unit 5 Lesson 1 Projectile Motion Falling things, and rockets ‘n’ that…

  2. Projectile Motion If an object is launched at an initial angle of θ0 with the horizontal, the analysis is similar except that the initial velocity has a vertical component. Initial Vertical Velocity does NOT = ZERO t = 2 Vy /g (if Y = 0) Initial Horizontal Velocity does NOT = Zero

  3. Projectile Motion It can be understood by analyzing the horizontal and vertical motions separately. Initial Vertical Velocity = ZERO t = √{2y/g} (if Vy =0) Initial Horizontal Velocity = Zero

  4. Projectile Motion Equations Horizontal Component: X = Vx(t) Vx = Vi COS  Vertical Component: Y = Vy(t) –(1/2)gt2 Vy = Vi SIN  t = 2 Vy /g (if Y = 0) t = √{2y/g} (if Vy =0)

  5. Projectile Motion 2-D Lesson 1 • Homework– Glencoe Page 150-151#’s 1 - 5 ALL

  6. Unit 5 Lesson 2: Objectives: Do Now: How far above the ground will a ball be above the ground, if it is thrown from ground level @ 30.625 m/s straight up in 2.50 seconds? Use 9.81 m/s2 for gravity Be able to gather information / data solve for X and Y Component Velocities solve for Range {X – displacement}solve for Time of Flightsolve for Height {Y – Displacement} Y = Vi (t) – ½ gt2 Y = 30.625 (2.5) – ½ (9.81)(2.5)2 Y = 76.5625 – 30.625 Y = 45.9375 m = 45.9 m : Homework Lesson 2 – Page 164 – 165 #45,52,57,59

  7. Horizontal Component: X = Vx(t) Vx = Vi COS  Vertical Component: Y = Vy(t) –(1/2)gt2 Vy = Vi SIN  t = 2 Vy /g (if Y = 0) t = √{2y/g} (if Vy =0) Review - Projectile Motion Equations

  8. Projectile Motion Projectile motion follows a parabolic path i.e. http://phet.colorado.edu/sims/projectile-motion/projectile-motion_en.html It has a horizontal (x) component and a vertical (y) component. The horizontal component is not affected by gravity, so use separate equationsfor the different components.

  9. Continued… What does this mean for objects dropped from the same height? Objects dropped from the same height will hit the ground at the same time, regardless of weight (ignoring differences in air resistance) http://www.animations.physics.unsw.edu.au/jw/projectiles.htm#1 Solve the same way you would any other kinematics problem… AICR State what you know, what do you want to find? Choose an equation and solve for your unknown. Look out for key words like.. “dropped”, “from rest” and “stationary:”

  10. Solving Problems Involving AICR Projectile Motion • Read the problem carefully, and choose the object(s) you are going to analyze. • Draw a diagram. • Choose an origin and a coordinate system. • Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g. • Examine the x and y motions separately. • 6. List known and unknown quantities. Remember that vx never changes, and that vy = 0 at the highest point. • 7. Plan how you will proceed. Use the appropriate equations; you may have to combine some of them.

  11. Objectives: Gain proficiency in collecting Data to solve for 2-D motion variables: Projectile Motion Lesson 3 Do NOW: How far down range will a cannonball go if fired at an angle of 30.0 deg an initial velocity of 18.5 m/s? assume no air resistance. Homework – Page: 164 - 165 #’s 42, 45, 51,56 Lab Set up

  12. What will hit the ground first a bullet dropped or a bullet fired from the same height?Discuss with your neighbor…http://www.youtube.com/watch?v=D9wQVIEdKh8

  13. Objects Under the Influence of Gravity:

  14. Projectile Motion The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g. This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.

  15. Projectile Motion Day TwoProblemsProjectile Motion Water slideWater Slide 2Giant House Slide

  16. Driving off a cliff!! How fast must the motorcycle leave the cliff to land at x = 90 m, y = -50 m? vx0 = ? y is positive upward y0 = 0 at top vy0 = 0 vx = vx0 = ? vy = -gt x = vx0t y = - (½)gt2 Time to Bottom: t = √2y/(-g) = 3.19 s vx0 = (x/t) = 28.2 m/s

  17. vx0= v0cos(θ0) = 16.0 15.97 m/s vy0= v0sin(θ0) = 12.036m/s y = y0+ vy0 t – (½)g t2 t= 2.531 sec (quadratic formula) x = vx0 t x = 40.540.49 m t/2 = 1.27 sec Ymax= y0+ vy0 t – (½)g t2= 1+ 12.0(1.27 ) – 4.9(1.27 )2 Y max = 8.3926 m Hang Time:2.53 seconds Range: 40.5 m Max Height: 8.39 m A punt!Find Hang time:Range:Maximum Height: v0 = 20.0 m/s, θ0 = 37.0º

  18. Test Review Unit 5 Motion in One-Dimension Displacement / Instantaneous Velocity / Acceleration Motion Diagrams Free Falling Objects / 1-Dimensional with constant Acceleration Motion in Two-Dimensions Vectors Coordinate System / Properties of Vectors Components of Vectors / Unit Vectors Adding Subtracting Multiplying Vectors (Graphically and Analytically) Scalar Products - Dot Products – (Honors: Cross Products) Motion in Two-Dimensions Projectile Range / Velocity Components / Maximum Height Time of Flight Circular Motion

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